Q.If A=[213−4] and B=[1−1−23], then verify that (AB)−1=B−1A−1. OR Prove that y+kyyyy+kyyyy+k=k2(3y+k)
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Inverse of a Product
Inverse of a Product: The "Socks and Shoes" Principle
You put on your socks first, then your shoes. To take them off, you can't remove the socks while the shoes are still on — you must reverse the order: shoes off first, then socks.
That's exactly the inverse of a product of matrices. If you apply transformation A first, then B, the combined effect is BA (read right-to-left: A acts first, then B). To undo it, undo B first, then A:
(AB)−1=B−1A−1
The order flips — forced by the logic of undoing.
Why the order must reverse
Check that B−1A−1 is the inverse of AB. We need (AB)(B−1A−1)=I and (B−1A−1)(AB)=I:
(AB)(B−1A−1)=A(BB−1)A−1=AIA−1=AA−1=I
B and B−1 cancel first, leaving A and A−1 to cancel. The other check works the same way:
(B−1A−1)(AB)=B−1(A−1A)B=B−1IB=B−1B=I
If you tried (AB)−1=A−1B−1 instead:
(AB)(A−1B−1)=A(BA−1)B−1
and BA−1 is not I — the matrices are in the wrong order. So the reversal is essential.
A common mistake is writing (AB)−1=A−1B−1. This is false unless A and B commute (which they almost never do). Always flip the order.
A concrete example with numbers
Let A=(1021) and B=(1101), with inverses:
A−1=(10−21),B−1=(1−101)
Then:
AB=(1021)(1101)=(3121),(AB)−1=(1−1−23)
Now compute B−1A−1:
B−1A−1=(1−101)(10−21)=(1−1−23)
They match. Try A−1B−1 and you'll get a different matrix — the wrong answer.
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Computing AB, then inverting it, and separately computing B−1A−1 confirms the standard reversal-order rule for the inverse of a matrix product; the OR part proves a determinant identity using row and column operations. …
(AB)−1=B−1A−1=[14/115/115/111/11] — verified.
A=[213−4], B=[1−1−23].
Step 1 — Compute AB:
AB=[2(1)+3(−1)1(1)+(−4)(−1)2(−2)+3(3)1(−2)+(−4)(3)]=[−155−14].
det(AB)=(−1)(−14)−5(5)=14−25=−11.
(AB)−1=−111[−14−5−5−1]=[14/115/115/111/11].
Step 2 — Compute A−1 and B−1:
detA=2(−4)−3(1)=−11, so A−1=−111[−4−1−32]=[4/111/113/11−2/11].
detB=1(3)−(−2)(−1)=3−2=1, so B−1=[3121].
Step 3 — Compute B−1A−1:
B−1A−1=[3121][4/111/113/11−2/11]=[1112+2114+1119−4113−2]=[14/115/115/111/11].
This matches (AB)−1 exactly, verifying (AB)−1=B−1A−1.
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- CBSE 2026Set 65/2/11 markMCQQ.For a square matrix A, (3A)−1= (A) 3A−1 (B) 9A−1 (C) 31A−1 (D) 91A−1
›Reveal solutionSolution
The inverse of a scalar multiple of a matrix, (kA)−1, is equal to k1A−1. For (3A)−1, this means the result is 31A−1.
Concept and Intuition
The inverse of a square matrix M, denoted M−1, is defined such that when M is multiplied by M−1, the result is the identity matrix I. That is, MM−1=M−1M=I. The identity matrix acts like the number 1 in scalar multiplication: MI=IM=M.
When we consider a scalar multiple of a matrix, say kA, we are essentially scaling every element of the matrix A by the scalar k. If we want to find the inverse of this new matrix (kA), we need to find a matrix that, when multiplied by kA, yields the identity matrix I.
Intuitively, if A−1 "undoes" the operation of A, and k "scales" A, then to "undo" kA, we would need to "un-scale" by k1 and then "un-matrix" by A−1. This suggests that the inverse of kA should involve k1 and A−1.
Let's verify this intuition formally.
Step-by-Step Derivation
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Recall the definition of an inverse matrix:
For any invertible square matrix M, its inverse M−1 satisfies the property MM−1=I, where I is the identity matrix of the same dimension.
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Apply the definition to (3A):
We are looking for (3A)−1. Let's denote this unknown inverse as X. By definition, X must satisfy:
(3A)X=I
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Propose a form for X based on intuition:
As discussed in the concept section, we expect X to be of the form cA−1 for some scalar c. Let's substitute this into the equation:
(3A)(cA−1)=I
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Use properties of scalar and matrix multiplication:
For any scalars k1,k2 and matrices M1,M2, we know that (k1M1)(k2M2)=(k1k2)(M1M2). Applying this property:
(3c)(AA−1)=I
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Substitute AA−1=I: …
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- CBSE 2022Set TERM11 markMCQQ.If A and B are invertible matrices then(a) (AB)⁻¹ = B⁻¹A⁻¹(b) (AB)⁻¹ = A⁻¹B⁻¹(c) (AB)⁻¹ = (BA)⁻¹(d) None of these
›Reveal solutionSolution
The inverse of a product reverses the order of the factors.
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